4 Answers
4

I believe you can find the value of $(3|p)$ without finding $p\equiv 5\pmod{12}$, by use of the second supplement to quadratic reciprocity. Recall that
$$
\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}.
$$
Since $p\equiv 2\pmod{3}$, you then have $(p|3)=(2|3)=-1$. So altogether,
$$
\left(\frac{3}{p}\right)\left(\frac{p}{3}\right)=\left(\frac{3}{p}\right)\left(\frac{2}{3}\right)=\left(\frac{3}{p}\right)(-1)=(-1)^{(p-1)(3-1)/4}=(-1)^{(p-1)/2}=1
$$
This implies $(3|p)=-1$.

You have an incorrect statement of the main case of quadratic reciprocity, for $p$ and $q$ odd primes. Instead of an equal sign between $(\frac p q)$ and $(\frac q p)$ there should be no symbol, or perhaps $\times$. The product $(\frac p q) (\frac q p)$ equals $(-1)^{\frac {p-1}{2} \cdot \frac {q-1}{2}}$, which is to say the product is $1$ except when $p$ and $q$ are both $3 \mod 4$, in which case it is $-1$.